Download Engr 6002: Ship Structures

Engr 6002: Ship Structures
Beam Review Problems &
Supplementary Information
Deformation of a Beam Under Transverse Loading
• Relationship between bending moment and
curvature for pure bending remains valid for
general transverse loadings.
1


M ( x)
EI
• Cantilever beam subjected to concentrated
load at the free end,
1


Px
EI
• Curvature varies linearly with x
1
• At the free end A, ρ  0,
A
• At the support B,
1
B
ρA  
 0,  B 
EI
PL
Deformation of a Beam Under Transverse Loading
• Overhanging beam
• Reactions at A and C
• Bending moment diagram
• Curvature is zero at points where the bending
moment is zero, i.e., at each end and at E.
1


M ( x)
EI
• Beam is concave upwards where the bending
moment is positive and concave downwards
where it is negative.
• Maximum curvature occurs where the moment
magnitude is a maximum.
• An equation for the beam shape or elastic curve
is required to determine maximum deflection
and slope.
Equation of the Elastic Curve
• From elementary calculus, simplified for beam
parameters,
d2y
1


dx 2
2 3 2
  dy 
1    
  dx  

d2y
dx 2
• Substituting and integrating,
EI
1

 EI
d2y
dx
2
 M x
x
dy
EI   EI
 M  x dx  C1
dx 
0
x
x
0
0
EI y   dx  M  x  dx  C1x  C2
Equation of the Elastic Curve
• Constants are determined from boundary
conditions
x
x
0
0
EI y   dx  M  x  dx  C1x  C2
• Three cases for statically determinant beams,
– Simply supported beam
y A  0,
yB  0
– Overhanging beam
y A  0, yB  0
– Cantilever beam
y A  0,  A  0
• More complicated loadings require multiple
integrals and application of requirement for
continuity of displacement and slope.
Direct Determination of the Elastic Curve From the
Load Distribution
• For a beam subjected to a distributed load,
d 2M
dM
 V x
dx
dV

  w x 
2
dx
dx
• Equation for beam displacement becomes
d 2M
dx
2
 EI
d4y
dx
4
  w x 
• Integrating four times yields
EI y  x     dx  dx  dx  w x dx
 16 C1x3  12 C2 x 2  C3 x  C4
• Constants are determined from boundary
conditions.
Statically Indeterminate Beams
• Consider beam with fixed support at A and roller
support at B.
• From free-body diagram, note that there are four
unknown reaction components.
• Conditions for static equilibrium yield
 Fx  0  Fy  0  M A  0
The beam is statically indeterminate.
• Also have the beam deflection equation,
x
x
0
0
EI y   dx  M  x  dx  C1x  C2
which introduces two unknowns but provides
three additional equations from the boundary
conditions:
At x  0,   0 y  0
At x  L, y  0
Sample Problem 9.1
SOLUTION:
• Develop an expression for M(x)
and derive differential equation for
elastic curve.
W 14  68
I  723 in 4
E  29  106 psi
P  50 kips
L  15 ft
a  4 ft
• Integrate differential equation twice
and apply boundary conditions to
obtain elastic curve.
For portion AB of the overhanging beam,
• Locate point of zero slope or point
(a) derive the equation for the elastic curve,
of maximum deflection.
(b) determine the maximum deflection,
• Evaluate corresponding maximum
(c) evaluate ymax.
deflection.
Sample Problem 9.1
SOLUTION:
• Develop an expression for M(x) and derive
differential equation for elastic curve.
- Reactions:
RA 
Pa
 a
 RB  P1   
L
 L
- From the free-body diagram for section AD,
M  P
a
x
L
0  x  L 
- The differential equation for the elastic
curve,
EI
d2y
a


P
x
2
L
dx
Sample Problem 9.1
• Integrate differential equation twice and apply
boundary conditions to obtain elastic curve.
EI
dy
1 a
  P x 2  C1
dx
2 L
1 a
EI y   P x3  C1x  C2
6 L
2
EI
d y
a


P
x
2
L
dx
at x  0, y  0 : C2  0
1 a
1
at x  L, y  0 : 0   P L3  C1L C1  PaL
6 L
6
Substituting,
dy
1 a
1
EI
  P x 2  PaL
dx
2 L
6
1 a
1
EI y   P x3  PaLx
6 L
6
2
dy PaL 
 x 

1  3  
dx 6 EI 
 L  
PaL2  x  x 
y
  
6 EI  L  L 
3


Sample Problem 9.1
• Locate point of zero slope or point
of maximum deflection.
2
dy
PaL 
 xm  
0
1  3  
dx
6 EI 
 L  
PaL2  x  x 
y
  
6 EI  L  L 
3


xm 
L
 0.577 L
3
• Evaluate corresponding maximum
deflection.

PaL2
ymax 
0.577  0.577 3
6 EI

PaL2
ymax  0.0642
6 EI
ymax

50 kips 48 in 180 in 2
 0.0642


6 29  106 psi 723 in 4

ymax  0.238in
Sample Problem 9.3
SOLUTION:
• Develop the differential equation for
the elastic curve (will be functionally
dependent on the reaction at A).
For the uniform beam, determine the
reaction at A, derive the equation for
the elastic curve, and determine the
slope at A. (Note that the beam is
statically indeterminate to the first
degree)
• Integrate twice and apply boundary
conditions to solve for reaction at A
and to obtain the elastic curve.
• Evaluate the slope at A.
Sample Problem 9.3
• Consider moment acting at section D,
MD  0
1  w0 x 2  x
RA x 
M 0
2  L  3
w0 x3
M  RA x 
6L
• The differential equation for the elastic
curve,
d2y
w0 x3
EI 2  M  RA x 
6L
dx
Sample Problem 9.3
• Integrate twice
4
dy
1
2 w0 x
EI
 EI  R A x 
 C1
dx
2
24 L
5
1
3 w0 x
EI y  R A x 
 C1x  C2
6
120 L
2
EI
d y
w0 x

M

R
x

A
6L
dx 2
3
• Apply boundary conditions:
at x  0, y  0 : C2  0
3
1
2 w0 L
at x  L,   0 :
RA L 
 C1  0
2
24
4
1
3 w0 L
at x  L, y  0 :
RA L 
 C1L  C2  0
6
120
• Solve for reaction at A
1
1
RA L3  w0 L4  0
3
30
RA 
1
w0 L 
10
Sample Problem 9.3
• Substitute for C1, C2, and RA in the
elastic curve equation,
5
1 1
 3 w0 x  1

EI y   w0 L  x 

w0 L3  x
6  10
120 L  120


y

w0
 x5  2 L2 x3  L4 x
120 EIL
• Differentiate once to find the slope,


dy
w0

 5 x 4  6 L2 x 2  L4
dx 120 EIL
at x = 0,
w0 L3
A 
120EI


Method of Superposition
Principle of Superposition:
• Deformations of beams subjected to
combinations of loadings may be
obtained as the linear combination of
the deformations from the individual
loadings
• Procedure is facilitated by tables of
solutions for common types of
loadings and supports.
Sample Problem 9.7
For the beam and loading shown,
determine the slope and deflection at
point B.
SOLUTION:
Superpose the deformations due to Loading I and Loading II as shown.
Sample Problem 9.7
Loading I
wL3
 B I  
6 EI
wL4
 yB I  
8EI
Loading II
wL3
C II 
48EI
wL4
 yC II 
128EI
In beam segment CB, the bending moment is
zero and the elastic curve is a straight line.
wL3
 B II  C II 
48EI
wL4
wL3  L  7 wL4
 yB II 

 
128 EI 48 EI  2  384 EI
Sample Problem 9.7
Combine the two solutions,
wL3 wL3
 B   B I   B II  

6 EI 48EI
7 wL3
B 
48EI
wL4 7 wL4
yB   yB I   yB II  

8EI 384 EI
41wL4
yB 
384EI
Application of Superposition to Statically
Indeterminate Beams
• Method of superposition may be
• Determine the beam deformation
applied to determine the reactions at
without the redundant support.
the supports of statically indeterminate
• Treat the redundant reaction as an
beams.
unknown load which, together with
• Designate one of the reactions as
the other loads, must produce
redundant and eliminate or modify
deformations compatible with the
the support.
original supports.
Sample Problem 9.8
For the uniform beam and loading shown,
determine the reaction at each support and
the slope at end A.
SOLUTION:
• Release the “redundant” support at B, and find deformation.
• Apply reaction at B as an unknown load to force zero displacement at B.
Sample Problem 9.8
• Distributed Loading:
4
3

w  2 
2 
3 2 
 yB w  
 L   2 L L   L  L 
24 EI  3 
3 
 3 
wL4
 0.01132
EI
• Redundant Reaction Loading:
2
2
RB  2   L 
RB L3
 yB R 
 L     0.01646
3EIL  3   3 
EI
• For compatibility with original supports, yB = 0
wL4
RB L3
0   yB w   yB R  0.01132
 0.01646
EI
EI
RB  0.688wL 
• From statics,
RA  0.271wL 
RC  0.0413wL 
Sample Problem 9.8
Slope at end A,
wL3
wL3
 A w  
 0.04167
24 EI
EI
2
0.0688wL  L   2  L  
wL3
 A R 
   L      0.03398
6 EIL  3  
EI
 3  
wL3
wL3
 A   A w   A R  0.04167
 0.03398
EI
EI
wL3
 A  0.00769
EI